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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 3"

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==Problem==
  
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Let <math>ABCD</math> be a parallelogram such that <math>E</math> is a point on <math>CD</math> such that <math>\frac{CE}{DE}=\frac{2}{3}.</math> Suppose that <math>BE</math> and <math>AC</math> intersect at <math>F.</math> If the area of triangle <math>AEF</math> is <math>15,</math> find the area of <math>ABCD</math>.
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==Solution==
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We have
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<cmath>\begin{align*}
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[AEF] &= [AEC]-[CEF]\\
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&= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\
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&= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\
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&= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\
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&= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\
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&= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\
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&= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\
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[ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}.
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\end{align*}</cmath>
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~pinkpig

Latest revision as of 17:18, 13 September 2025

Problem

Let $ABCD$ be a parallelogram such that $E$ is a point on $CD$ such that $\frac{CE}{DE}=\frac{2}{3}.$ Suppose that $BE$ and $AC$ intersect at $F.$ If the area of triangle $AEF$ is $15,$ find the area of $ABCD$.

Solution

We have \begin{align*} [AEF] &= [AEC]-[CEF]\\ &= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\ &= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\ &= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\ &= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\ &= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\ &= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\ [ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}. \end{align*}

~pinkpig

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